Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<pair<string, int> > v1, v2;
  int mat1[101][101];
  int mat2[101][101];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> mat1[i][j];
      mat2[i][j] = mat1[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    int maxi = 1e9;
    for (int j = 0; j < m; j++) {
      maxi = min(maxi, mat1[i][j]);
    }
    for (int j = 0; j < m; j++) mat1[i][j] -= maxi;
    for (int j = 0; j < maxi; j++) v1.push_back(make_pair("row", i + 1));
  }
  for (int j = 0; j < m; j++) {
    int maxi = 1e9;
    for (int i = 0; i < n; i++) {
      maxi = min(maxi, mat1[i][j]);
    }
    for (int i = 0; i < n; i++) mat1[i][j] -= maxi;
    for (int i = 0; i < maxi; i++) v1.push_back(make_pair("col", j + 1));
  }
  for (int j = 0; j < m; j++) {
    int maxi = 1e9;
    for (int i = 0; i < n; i++) {
      maxi = min(maxi, mat2[i][j]);
    }
    for (int i = 0; i < n; i++) mat2[i][j] -= maxi;
    for (int i = 0; i < maxi; i++) v2.push_back(make_pair("col", j + 1));
  }
  for (int i = 0; i < n; i++) {
    int maxi = 1e9;
    for (int j = 0; j < m; j++) {
      maxi = min(maxi, mat2[i][j]);
    }
    for (int j = 0; j < m; j++) mat2[i][j] -= maxi;
    for (int j = 0; j < maxi; j++) v2.push_back(make_pair("row", i + 1));
  }
  bool c1 = true, c2 = true;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (mat1[i][j] != 0) c1 = false;
      if (mat2[i][j] != 0) c2 = false;
    }
  }
  if (c1 && c2) {
    if (v1.size() < v2.size()) {
      cout << v1.size() << endl;
      for (int i = 0; i < v1.size(); i++)
        cout << v1[i].first << " " << v1[i].second << endl;
    } else {
      cout << v2.size() << endl;
      for (int i = 0; i < v2.size(); i++)
        cout << v2[i].first << " " << v2[i].second << endl;
    }
  } else if (c1) {
    cout << v1.size() << endl;
    for (int i = 0; i < v1.size(); i++)
      cout << v1[i].first << " " << v1[i].second << endl;
  } else if (c2) {
    cout << v2.size() << endl;
    for (int i = 0; i < v2.size(); i++)
      cout << v2[i].first << " " << v2[i].second << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}